If `A(theta)=[(sin theta, i cos theta),(i cos theta, sin theta)]`, then which of the following is not true ?

Options

A. `A(theta)^(-1)=A(pi-theta)`
B. `A(theta)+A(pi+theta)` is a null matrix
C. `A(theta)` is invertible for all `theta in R`
D. `A(theta)^(-1)=A(-theta)`

Explanation

To solve the problem, we need to analyze the given matrix \( A(\theta) = \begin{pmatrix} \sin \theta & i \cos \theta \\ i \cos \theta & \sin \theta \end{pmatrix} \) and determine which of the provided statements about it is not true.

### Step 1: Calculate the Determinant of \( A(\theta) \)

The determinant of a 2x2 matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by the formula:
\[ \text{det}(A) = ad - bc \]
For our matrix \( A(\theta) \):
- \( a = \sin \theta \)
- \( b = i \cos \theta \)
- \( c = i \cos \theta \)
- \( d = \sin \theta \)

Thus, the determinant is:
\[ \text{det}(A(\theta)) = \sin \theta \cdot \sin \theta - (i \cos \theta)(i \cos \theta) \]
Calculating this gives:
\[ \text{det}(A(\theta)) = \sin^2 \theta - (-1) \cos^2 \theta = \sin^2 \theta + \cos^2 \theta \]
Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \text{det}(A(\theta)) = 1 \]

### Step 2: Check if \( A(\theta) \) is Invertible

Since the determinant is 1 (which is not equal to zero), the matrix \( A(\theta) \) is invertible for all \( \theta \).

### Step 3: Check the Statement \( A(\theta)^{-1} = A(\pi - \theta) \)

To find \( A(\pi - \theta) \):
\[ A(\pi - \theta) = \begin{pmatrix} \sin(\pi - \theta) & i \cos(\pi - \theta) \\ i \cos(\pi - \theta) & \sin(\pi - \theta) \end{pmatrix} \]
Using the identities \( \sin(\pi - \theta) = \sin \theta \) and \( \cos(\pi - \theta) = -\cos \theta \):
\[ A(\pi - \theta) = \begin{pmatrix} \sin \theta & -i \cos \theta \\ -i \cos \theta & \sin \theta \end{pmatrix} \]
Next, we calculate \( A(\theta)^{-1} \). The inverse of a 2x2 matrix \( A \) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]
Thus,
\[ A(\theta)^{-1} = \begin{pmatrix} \sin \theta & -i \cos \theta \\ -i \cos \theta & \sin \theta \end{pmatrix} \]
This shows that:
\[ A(\theta)^{-1} = A(\pi - \theta) \]
So, the statement \( A(\theta)^{-1} = A(\pi - \theta) \) is true.

### Step 4: Check the Statement \( A(\pi + \theta) = -A(\theta) \)

For \( A(\pi + \theta) \):
\[ A(\pi + \theta) = \begin{pmatrix} \sin(\pi + \theta) & i \cos(\pi + \theta) \\ i \cos(\pi + \theta) & \sin(\pi + \theta) \end{pmatrix} \]
Using the identities \( \sin(\pi + \theta) = -\sin \theta \) and \( \cos(\pi + \theta) = -\cos \theta \):
\[ A(\pi + \theta) = \begin{pmatrix} -\sin \theta & -i \cos \theta \\ -i \cos \theta & -\sin \theta \end{pmatrix} = -A(\theta) \]
Thus, the statement \( A(\pi + \theta) = -A(\theta) \) is true.

### Step 5: Check the Statement \( A(\theta)^{-1} = A(-\theta) \)

For \( A(-\theta) \):
\[ A(-\theta) = \begin{pmatrix} \sin(-\theta) & i \cos(-\theta) \\ i \cos(-\theta) & \sin(-\theta) \end{pmatrix} \]
Using the identities \( \sin(-\theta) = -\sin \theta \) and \( \cos(-\theta) = \cos \theta \):
\[ A(-\theta) = \begin{pmatrix} -\sin \theta & i \cos \theta \\ i \cos \theta & -\sin \theta \end{pmatrix} \]
This is not equal to \( A(\theta)^{-1} \) which is:
\[ A(\theta)^{-1} = \begin{pmatrix} \sin \theta & -i \cos \theta \\ -i \cos \theta & \sin \theta \end{pmatrix} \]
Thus, the statement \( A(\theta)^{-1} = A(-\theta) \) is not true.

### Conclusion

The statement that is not true is:
**Option 4: \( A(\theta)^{-1} = A(-\theta) \)**.